Question

The circle $x^2+y^2-8x=0$ and hyperbola $\frac{x^2}{9}-\frac{y^2}{4}=1$ intersect at the points $A$ and $B$
Equation of the circle with $AB$ as its diameter is

• A. (A) $x^2+y^2-12x+24=0$
• B. (B) $x^2+y^2+12x+24=0$
• C. (C) $x^2+y^2+24x-12=0$
• D. (D) $x^2+y^2-24x-12=0$

Answer 1

The correct option is D (D) $x^2+y^2-24x-12=0$

The equation of circle is given as,

$x^2+y^2-8x=0$

$\therefore (x^2-8x)+y^2=0$

$\therefore (x^2-8x+16)-16+{(y-0)}^2=0$

$\therefore {(x-4)}^2+{(y-0)}^2=16$

$\therefore {(x-4)}^2+{(y-0)}^2={\left(4\right)}^2$

Comparing this equation with standard form i.e. ${(x-h)}^2+{(y-k)}^2={\left(r\right)}^2$, we get

coordinates of center = $C(4,0)$ and radius $r=4$

Thus, center of circle lies on right of origin and circle touches y axis at origin as shown in figure.

Now, Equation of hyperbola is,

$\frac{x^2}{9}-\frac{y^2}{4}=1$

$\therefore \frac{y^2}{4}=\frac{x^2}{9}-1$

Multiply LHS and RHS by 4, we get,

$y^2=\frac{4x^2}{9}-4$

$\therefore y=\sqrt{\frac{4x^2}{9}-4}$

When $x\to \infty$, we can neglect 4 in the bracket.

$\therefore y=\sqrt{\frac{4}{9}{x}^2}$

$\therefore y=\pm \frac{2}{3}x$

Thus, we get two lines extending up to infinity as shown in figure.

Now, we have to find point of intersection of hyperbola with x axis. This can be done by equating y to zero in equation of hyperbola

$\therefore \sqrt{\frac{4}{9}{x}^2-4}=0$

Squaring both sides, we get,

$\therefore \frac{4}{9}{x}^2-4=0$

$\therefore \frac{4}{9}{x}^2=4$

$\therefore x^2=9$

$\therefore x=\pm 3$

Thus, hyperbola will pass through $x=3$ and $x=-3$ as shown in figure.

To find point of intersection of hyperbola and circle-

Point of intersection of hyperbola and circle is common in both the curves. Thus, it must satisfy equation of hyperbola and circle.

From equation of circle, we can write,

$x^2+y^2-8x=0$

$\therefore y^2=8x-x^2$

Put this value in equation of hyperbola, we get,

$\therefore \frac{x^2}{9}-\frac{8x-x^2}{4}=1$

$\therefore \frac{x^2}{9}-2x+\frac{x^2}{4}=1$

Taking LCM of denominators which is 36. Thus, converting each denominator to 36, we get,

$\therefore \frac{4x^2}{36}-\frac{72x}{36}+\frac{9x^2}{36}=1$

$\therefore 4x^2-72x+9x^2=36$

$\therefore 13x^2-72x-36=0$

This is a quadratic equation in x. Thus, using formula method to find roots of the equation.

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Here, $a=13$, $b=-72$, $c=-36$

$\therefore x=\frac{-(-72)\pm \sqrt{{(-72)}^2-4\times 13\times (-36)}}{2\times 13}$

$\therefore x=\frac{72\pm \sqrt{(72\times 72)+(4\times 13\times 36)}}{26}$

$\therefore x=\frac{72\pm \sqrt{(2\times 36\times 2\times 36)+(4\times 13\times 36)}}{26}$

$\therefore x=\frac{72\pm \sqrt{(4\times 36\times 36)+(4\times 36\times 13)}}{26}$

$\therefore x=\frac{72\pm \sqrt{(4\times 36)(36+13)}}{26}$

$\therefore x=\frac{72\pm \sqrt{4\times 36\times 49}}{26}$

$\therefore x=\frac{72\pm (2\times 6\times 7)}{26}$

$\therefore x=\frac{72\pm 84}{26}$

$\therefore x=\frac{72+84}{26}$ or $x=\frac{72-84}{26}$

$\therefore x=6$ or $x=\frac{-6}{13}$

As point of intersection lies to the right of origin, we can neglect negative root of x.

$\therefore x=6$

Put this value of x in equation of circle, we get,

$y^2=8\left(6\right)-{\left(6\right)}^2$

$y^2=48-36=12$

$\therefore y=\pm \sqrt{12}$

Thus, coordinates of point of intersection are,

$A(6,\sqrt{12})$ and $B(6,-\sqrt{12})$

Now, we have to find equation of circle having AB as diameter.

From the figure, it is clear that center of circle is $C(6,0)$ and radius of circle is $r=\sqrt{12}$

Thus, equation of circle is,

${(x-6)}^2+{(y-0)}^2={\left(\sqrt{12}\right)}^2$

$\therefore x^2-12x+36+y^2=12$

$\therefore x^2+y^2-12x+36-12=0$

$\therefore x^2+y^2-12x+24=0$

Thus, answer is option (A)